Out of these we see that there are 3 occasions where there is exactly one heads.
If we let S be the sample space of all possable outcomes, and let E be the event of having exactly one heads, then a basic rule of probability indicates that the probability P(E)=|E|/|S|=3/7.

However, using another method:
fix one coin throw's result as heads. Then the probability of the other two throws being tails is (1/2)(1/2)=1/4.
Since we can fix heads 3 times, this indicates that the probability is 3(1/4)=3/4.

Can anyone spot where the flaw is in either of these attempts at a solution?
thanks
(this isn't a hw question btw)